2011S midterm 1 - key

# 2011S midterm 1 - key - Name X.500 ID Midterm 1 Answer Key...

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Name: X.500 ID: Midterm 1 – Answer Key 100 points CSci 4511, Thursday March 3, 2011 Closed book, notes, laptop, cell phone, calculator. 1 page of notes Unless the question specifically asks for more detail, you may use short answers. 1. 4 points State the solution path and cost found by breadth-first search ( not uniform cost search). S Æ A Æ D Æ G Cost: 3 Costs in breadth-first search is the number of steps i think some students confused “solution path” with “what nodes will it search” 2. 4 points In what sense is IDA* preferable to A*? Uses less memory Several people mentioned that it’s interruptible. i don’t think that’s what Korff, the guy who invented it, had in mind when he made it, but it was obviously a big deal in the chess problem so that answer was also worth full credit 3. 6 points Name one reason why you would use depth-first search. Name a reason why you would not. Pro: Uses less memory, good choice for short, fat problems Con: Not optimal, gets caught on loops, does not work on problems with infinite depth, wrong choice for tall skinny problems A lot of people brought up variable assignment problems. If you said it finds answers faster than BFS in the best case and the same as BFS in the worst, you got full credit. If you said it works better than BFS for those problems but don’t say why, you lost points. If you said it works on problems where the solution depth is know (not “it works better”, just “it works”), you lost a lot of points 4. 4 points You are at a node where g(n)=10 and h(n)=10. This node is on the optimal path. What is the actual cost (C*) of the path through that node to the solution? If this is a trick question, say why. S A B C 10 1 D E F G H 1 20 8 5 12 6 4 3

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We have no idea what the actual cost is. All we know is that it’s at least 20, but since h(n) can be an underestimate and g(n) is perfect, f(n) can also be an underestimate 5. 1 point each Are these heuristics admissible (T/F)? Note: C(n, goal) is the true cost from the node to the goal. a. h1(n)=0 T b. h2(n)=g(n) F c. h3(n)= random(0, C(n, goal)) T d. h4(n)=2C(n, goal) F e. h5(n)=½C(n, goal) T Most people got this with the exception of 5b, g(n). As you get closer to the solution, g grows and h shrinks. If you used g(n) as your heuristic, h(n) gets bigger as you get closer to the goal, which is the wrong direction. Suppose you’re on a path C*=20. If you’re at the start of the path, you might have g(n)=5, h(n)=15. Near the end of the path, g(n)=15, h(n)=5. If you used h(n)=g(n) then near the end of the path the remaining cost is 5you’re your estimate is 15, 3 times the actual cost and clearly not admissible. And what happens with that inadmissible heuristic? You’d have f(n)=g(n)+h(n) = 15+15 = 30 which is > C*, so the solution would be below where it should be and a non-optimal solution might be found first. It also has the effect of pushing A* away from the goal, not towards it 6. 4 points The open list contains 101 nodes. One is the goal, f(goal)=45. Of the remaining 100 nodes, 10 have the cost f(n)=10, 10 have f(n)=20, 10 have f(n)=30, etc. all the way to 10 nodes with f(n)=100. Of these, only 8 nodes are on the optimal path. Including the goal node, how many of
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2011S midterm 1 - key - Name X.500 ID Midterm 1 Answer Key...

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